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Câu hỏi: Cho hàm số $f\left( x \right)$ thỏa mãn $\underset{0}{\overset{1}{\mathop \int }} \left( 2x+1 \right).{f}'\left( x \right)dx=20$ và $3f\left( 1 \right)-f\left( 0 \right)=4.$ Tích phân $\underset{0}{\overset{1}{\mathop \int }} f\left( x \right)dx$ bằng
A. -16.
B. 16.
C. -8
D. 8.
A. -16.
B. 16.
C. -8
D. 8.
Ta có $20=\int\limits_{0}^{1}{\left( 2\text{x}+1 \right).{f}'\left( x \right)d\text{x}=\int\limits_{0}^{1}{\left( 2\text{x}+1 \right)d\left[ f\left( x \right) \right]=\left( 2\text{x}+1 \right).f\left( x \right)\mathop{|}_{0}^{1}}}-\int\limits_{0}^{1}{f\left( x \right)d\left( 2\text{x}+1 \right)}$
$=3f\left( 1 \right)-f\left( 0 \right)-2\underset{0}{\overset{1}{\mathop \int }} f\left( x \right)dx=4-2\underset{0}{\overset{1}{\mathop \int }} f\left( x \right)dx\Rightarrow \underset{0}{\overset{1}{\mathop \int }} f\left( x \right)dx=-8.$ Chọn C.
$=3f\left( 1 \right)-f\left( 0 \right)-2\underset{0}{\overset{1}{\mathop \int }} f\left( x \right)dx=4-2\underset{0}{\overset{1}{\mathop \int }} f\left( x \right)dx\Rightarrow \underset{0}{\overset{1}{\mathop \int }} f\left( x \right)dx=-8.$ Chọn C.
Đáp án C.